$a$) $(x-2).(xy+3) = -5$
$⇒$ $x-2;xy+3$ $∈$ `Ư(5)={±1;±5}`
Ta có bảng:
$\left[\begin{array}{ccc}x-2&-5&-1&1&5\\xy+3&1&5&-5&-1\\x&-3&1&3&7\\y&KTM&2&KTM&KTM\end{array}\right]$
Vậy `(x;y)=(1;2)`
$b$)$- 2x + 3y + xy = 7$
$⇔ x(y - 2) + 3y = 7$
$⇔ x(y-2) + 3y - 6 = 1$
$⇔ x(y-2) + 3(y-2) = 1$
$⇔ (x+3)(y-2)=1$
$⇒$ $x+3;y-2$ $∈$ `Ư(1)={±1}`
Ta có bảng:
$\left[\begin{array}{ccc}x+3&-1&1\\y-2&-1&1\\x&-4&-2\\y&1&3\end{array}\right]$
Vậy `(x;y)=(-4;1);(-2;3)`
